Momentum and Collisions动量与碰撞

Velocity describes how one object moves, but it is not conserved when objects interact. Momentum is, because Newton's third law makes the forces between two interacting objects equal and opposite. When body A pushes B, B pushes back on A with the same force for the same time, so the impulses are equal and opposite — whatever momentum A gains, B loses. Summed over the pair, the total momentum cannot change. This is why momentum, weighted by mass, is the right bookkeeping variable for collisions and explosions.速度描述单个物体如何运动，但物体相互作用时它并不守恒。动量则守恒，因为牛顿第三定律使两个相互作用物体之间的力大小相等、方向相反。当物体 A 推 B 时，B 以相同的力、相同的时间反推 A，故两者的冲量等大反向 — A 获得多少动量，B 就失去多少。对这一对求和，总动量不可能改变。这正是为何以质量加权的动量是碰撞与爆炸的恰当记账变量。

Newton actually stated his second law in terms of momentum: the net force equals the rate of change of momentum, $\vec F_{\text{net}} = \frac{\Delta \vec p}{\Delta t}$. For constant mass this reduces to the familiar $\vec F = m\vec a$, but the momentum form is more general and is the direct route to the impulse-momentum theorem of §2.牛顿其实是用动量陈述其第二定律的：净力等于动量的变化率，$\vec F_{\text{net}} = \frac{\Delta \vec p}{\Delta t}$。质量恒定时它化为熟悉的 $\vec F = m\vec a$，但动量形式更普遍，也是通往 §2 冲量-动量定理的直接路径。

Real collision forces are not constant — they spike and fade over a few milliseconds. The general definition of impulse is the area under the force-time graph, $J = \int F\,dt$. At HS level you read this area off the graph the same way you read displacement off a velocity-time graph in the Kinematics unit.真实的碰撞力并非恒定 — 它在几毫秒内骤升又消退。冲量的普遍定义是力-时间图下的面积，$J = \int F\,dt$。在高中阶段，你像在运动学单元里从速度-时间图读位移那样从图上读出这块面积。

The average force $\bar F$ is the constant force that would give the same area (same impulse) over the same time: $\bar F\,\Delta t = J$. So when a problem asks for "the average force," it is asking you to flatten that spiky curve into the equivalent rectangle. This is the precise sense in which a longer, gentler contact (airbag) and a short, violent one (hard dashboard) can deliver the same impulse but wildly different peak forces.而平均力 $\bar F$ 是在相同时间内给出相同面积（相同冲量）的那个恒力：$\bar F\,\Delta t = J$。所以当题目问"平均力"时，是要你把那条尖峰曲线压平为等效的矩形。这正是为何更长更柔的接触（气囊）与短促剧烈的接触（硬仪表台）能施加相同冲量却有天差地别的峰值力。

Consider two objects A and B interacting (a collision, a push). During contact, A exerts a force $\vec F_{AB}$ on B, and by Newton's third law B exerts $\vec F_{BA} = -\vec F_{AB}$ on A. They act for the same time $\Delta t$, so the impulses are equal and opposite:考虑两个相互作用的物体 A 与 B（碰撞、推挤）。接触期间 A 对 B 施力 $\vec F_{AB}$，由牛顿第三定律，B 对 A 施力 $\vec F_{BA} = -\vec F_{AB}$。它们作用相同的时间 $\Delta t$，故冲量等大反向：

Adding them, $\Delta \vec p_A + \Delta \vec p_B = 0$, so the total momentum $\vec p_A + \vec p_B$ does not change. Any number of internal forces cancel the same way, in equal-and-opposite pairs. Only an external net force (one from outside the chosen system) can change the total. This is why choosing the system carefully — deciding which objects are "inside" — is the first move in every conservation problem.两式相加，$\Delta \vec p_A + \Delta \vec p_B = 0$，故总动量 $\vec p_A + \vec p_B$ 不变。任意多个内力都以等大反向的成对方式同样抵消。只有外部净力（来自所选系统之外）才能改变总动量。这正是为何谨慎选择系统——决定哪些物体在"内部"——是每道守恒问题的第一步。

Solving the momentum and kinetic-energy equations together for a 1D elastic collision (masses $m_1$, $m_2$; initial velocities $u_1$, $u_2$) yields a clean pair of formulas:对一维弹性碰撞（质量 $m_1$、$m_2$；初速度 $u_1$、$u_2$）联立动量与动能方程，得到一对简洁公式：

A more memorable equivalent is the relative-velocity reversal: in any 1D elastic collision the relative velocity of approach equals the relative velocity of separation,一个更好记的等价说法是相对速度反向：在任何一维弹性碰撞中，接近的相对速度等于分离的相对速度，

Setting $m_1 = m_2$ in the formulas gives $v_1 = u_2$, $v_2 = u_1$ (the velocity swap of Example 4). Setting $m_2 \gg m_1$ with $u_2 = 0$ gives $v_1 \approx -u_1$ (a light ball bounces back off a heavy wall at nearly its incoming speed). These limits are worth knowing for the AP and IB momentum units.在公式中令 $m_1 = m_2$ 得 $v_1 = u_2$、$v_2 = u_1$（即例 4 的速度交换）。令 $m_2 \gg m_1$ 且 $u_2 = 0$ 得 $v_1 \approx -u_1$（轻球以近乎来速从重墙弹回）。这些极限值得为 AP 与 IB 动量单元记住。

Momentum conservation fixes the total momentum $p_{\text{tot}}$ after the collision, but it does not by itself fix how the kinetic energy is shared. Of all the velocity outcomes allowed by a fixed total momentum, the one that minimises kinetic energy is the one where both objects move with the same velocity — that is, where they stick together. Any relative motion between the two objects after the collision carries extra kinetic energy, so removing all relative motion (sticking) removes the most.动量守恒固定了碰后总动量 $p_{\text{tot}}$，但它本身并不固定动能如何分配。在固定总动量允许的所有速度结果中，使动能最小的那个是两物体以相同速度运动的情形 — 也就是它们粘在一起。碰后两物体间的任何相对运动都携带额外动能，故消除全部相对运动（粘合）即消除最多。

The kinetic energy that cannot disappear is the part tied up in the centre-of-mass motion, $\tfrac{1}{2}(m_1+m_2)v_{\text{cm}}^2$, which momentum conservation locks in. The "extra" kinetic energy associated with motion relative to the centre of mass is what is available to be converted to heat and sound — and a completely inelastic collision converts all of it. This is why crash-safety design (NGSS HS-PS2-3) deliberately makes car collisions more inelastic, channelling kinetic energy into crumple zones instead of into the passengers.不可能消失的那部分动能是束缚于质心运动中的 $\tfrac{1}{2}(m_1+m_2)v_{\text{cm}}^2$，由动量守恒锁定。与质心相对运动相关的"额外"动能才是可被转化为热与声的部分 — 完全非弹性碰撞把它全部转化。这正是为何撞击安全设计（NGSS HS-PS2-3）刻意让车祸更非弹性，把动能导入溃缩区而非乘客。

A famous result for billiards: when a moving ball collides elastically with an identical stationary ball in two dimensions (a glancing, off-centre hit), the two balls always separate at $90^{\circ}$ to each other, provided the collision is elastic and the masses are equal.台球的一个著名结论：当一个运动的球在二维中与一个相同的静止球发生弹性碰撞（擦边、偏心击打）时，只要碰撞是弹性的且质量相等，两球总以彼此成 $90^{\circ}$ 的方向分开。

The proof combines both conservation laws. Momentum gives $\vec u_1 = \vec v_1 + \vec v_2$ (writing the initial velocity as the vector sum of the two finals). Squaring this vector equation gives $u_1^2 = v_1^2 + v_2^2 + 2\,\vec v_1 \cdot \vec v_2$. Kinetic-energy conservation for equal masses gives $u_1^2 = v_1^2 + v_2^2$. Subtracting, $2\,\vec v_1 \cdot \vec v_2 = 0$, so the dot product of the two final velocities is zero — they are perpendicular. This elegant interplay of momentum (a vector law) and energy (a scalar law) is exactly the relationship Ontario SPH4U C2 asks students to investigate, and it reappears in AP Physics and IB Physics HL collision problems.证明结合两条守恒律。动量给出 $\vec u_1 = \vec v_1 + \vec v_2$（把初速度写成两个末速度的矢量和）。将此矢量方程平方得 $u_1^2 = v_1^2 + v_2^2 + 2\,\vec v_1 \cdot \vec v_2$。等质量的动能守恒给出 $u_1^2 = v_1^2 + v_2^2$。相减得 $2\,\vec v_1 \cdot \vec v_2 = 0$，故两末速度的点积为零 — 它们相互垂直。这种动量（矢量律）与能量（标量律）的优雅相互作用正是安大略 SPH4U C2 要求学生探究的关系，也在 AP Physics 与 IB Physics HL 碰撞问题中再现。